Evaluate the amount of heat produced during a day by a `beta^(-)`- active `Na^(24)` preparation of mass `m= 1.0 mg`. The beta particles are assumed to possess an average kinetic energy equal to `1//3` of the highest possible energy of the given decay. The half-life of `Na^(24)` is `T= 15 hours`.

Explanation

The masses are
`Na^(24)= 24-0.00903 amu` and `Mg^(24)= 24-0.01496 amu`
The reaction is
`Na^(24) rarr Mg^(24)+e^(-)+bar(v)_(e )`
The maximum `K.E` of electrons is
`0.00593xx93MeV= 5.52MeV`
Average `K.E.` according to the problem is then `(5.52)/(3)= 1.84MeV`
The initial number of `Na^(24)` is
`(10^(-3)xx6.023xx10^(23))/(24)=2.51xx10^(19)`
The fraction decying in a day is
`1-(2)^(-24//15)= 0.67`
Hence the heat produced in a day is
`0.67xx2.51xx10^(19)xx1.84xx1.602xx10^(-13)` Joul `= 4.95MJ`